# Total moment of inertia of a two particle system

by Linus Pauling
Tags: inertia, moment, particle
 P: 190 1. Find the moment of inertia I_x of particle a with respect to the x axis (that is, if the x axis is the axis of rotation), the moment of inertia I_y of particle a with respect to the y axis, and the moment of inertia I_z of particle a with respect to the z axis (the axis that passes through the origin perpendicular to both the x and y axes). Particle a is located 3r from the y axis, and particle b is located r away. 2. I = mr^2 I = SUMM(m_i*r_i^2) 3. I_a = 9mr^2 I_b = mr^2 Total I = 10mr^2 No idea what I am doing wrong
Mentor
P: 40,278
 Quote by Linus Pauling Particle a is located 3r from the y axis, and particle b is located r away.
Give the complete coordinates and the mass of each particle.
 P: 190 Particle a: distance from y axis = 3r distance from x axis = r Particle b: distance from y axis = r distance from x axis = -4r It doesn't say anything explicitly about mass, and so I think we assume each particle has mass m.
Mentor
P: 40,278

## Total moment of inertia of a two particle system

 Quote by Linus Pauling 3. I_a = 9mr^2 I_b = mr^2 Total I = 10mr^2 No idea what I am doing wrong
Looks like you've found I_y. What about I_x?
 P: 190 I_x would be m(r)^2 + m(-4r)^2 = mr^2 + 16mr^2 = 17mr^2. Am I simply adding that to I_y to get I_total? It asks for total I with respect to the y axis, though, why would I need I_x?
 P: 190 Shoot, just realized I posted to wrong question. Here's what I am trying to answer: Find the total moment of inertia I of the system of two particles shown in the diagram with respect to the y axis
Mentor
P: 40,278
 Quote by Linus Pauling Here's what I am trying to answer: Find the total moment of inertia I of the system of two particles shown in the diagram with respect to the y axis
Assuming the diagram matches your description, your answer seems correct to me. (Post the diagram, if you can.)
 P: 190
 Mentor P: 40,278 Looks good to me. I say your answer is correct.
 P: 190 Ok, I requested the correct answer: 11mr^2 WTF?!
Mentor
P: 40,278
 Quote by Linus Pauling Ok, I requested the correct answer: 11mr^2
Sounds bogus to me. You might want to mention this to your instructor.
 P: 190 Ok, now: Using the formula for kinetic energy of a moving particle K=\frac{_1}{^2}mv^2, find the kinetic energy K_a of particle a and the kinetic energy K_b of particle b. Express your answers in terms of m, omega, and r separated by a comma. I know I need to calculate linear speed, how do I do that? I know it's v=omega*r but I don't see how to put this together.....
Mentor
P: 40,278
 Quote by Linus Pauling I know I need to calculate linear speed, how do I do that?
What's the relationship between tangential speed and angular speed for something going in a circle?
 P: 190 Just edited before you replied: v=omega*r But what are omega and r here?!
 P: 190 Particle A: Ka = 0.5m(3*omega*r)^2 Kb = 0.5m(omega*r)^2 Apparently Kb is wrong but I did it the same way as Ka.............
 P: 190 Ok I got Kb = m(omega*r)^2 What is total kinteic eneregy?
Mentor
P: 40,278
 Quote by Linus Pauling Ka = 0.5m(3*omega*r)^2 Kb = 0.5m(omega*r)^2 Apparently Kb is wrong but I did it the same way as Ka.............
These both look OK to me.

 Quote by Linus Pauling Ok I got Kb = m(omega*r)^2
How did you get that? (Are you posting this problem exactly as given, word for word?)

 What is total kinetic energy?
Just add up the kinetic energy of each mass.

 Related Discussions General Engineering 2 Introductory Physics Homework 1 Calculus & Beyond Homework 6 Introductory Physics Homework 0 Introductory Physics Homework 4